Find Derivative: Solve G'(3) Problem

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Discussion Overview

The discussion revolves around finding the derivative G'(3) based on the function G(x) defined in terms of f(x) and g(x). Participants are analyzing how to derive G'(3) using the derivatives of f and g, as well as values obtained from a graph.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • Some participants propose that G'(3) can be calculated using the formula G'(x) = 6f'(x) - g'(x), leading to G'(3) = 6f'(3) - g'(3).
  • One participant suggests that G'(3) might equal -26, but this is questioned by others.
  • Another participant claims to have determined values from the graph, stating f'(3) = 6 and g'(3) = 1.
  • Contrarily, a different participant argues that f'(3) should be -3 based on observed changes in the graph, while agreeing that g'(3) = 1.
  • Further clarification is sought regarding how the value of -3 for f'(3) was derived, with reference to the concept of rise over run.

Areas of Agreement / Disagreement

Participants express differing views on the values of f'(3) and g'(3), leading to multiple competing perspectives on the calculation of G'(3). No consensus is reached regarding the correct values to use.

Contextual Notes

Participants rely on graphical observations to determine the derivatives, which may involve assumptions about the accuracy of the graph and the interpretation of the slope.

Who May Find This Useful

Students or individuals interested in calculus, particularly in understanding derivatives and their applications in function analysis.

Teh
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View attachment 6132

Does it want me to plug in G'(3) into the equation?
 

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We are told:

$$G(x)=6f(x)-g(x)$$

And so we know:

$$G'(x)=6f'(x)-g'(x)$$

which means:

$$G'(3)=6f'(3)-g'(3)$$

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D
 
MarkFL said:
We are told:

$$G(x)=6f(x)-g(x)$$

And so we know:

$$G'(x)=6f'(x)-g'(x)$$

which means:

$$G'(3)=6f'(3)-g'(3)$$

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D
will it be -26 when G'(3)
 
the said:
will it be -26 when G'(3)

What values do you get from the graph for $f'(3)$ and $g'(3)$?
 
MarkFL said:
What values do you get from the graph for $f'(3)$ and $g'(3)$?

you get f'(3) = 6 and g'(3) = 1
 
Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.
 
MarkFL said:
Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.

how did you get -3 from f'(3)?
 
the said:
how did you get -3 from f'(3)?

Rise over run...rise is -3 when run is 1...:D
 

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